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Diano4ka-milaya [45]
3 years ago
10

If you remove the last digit (one’s place) from a 4-digit whole number, the resulting number is a factor of the 4-digit number.

How many such 4-digit numbers are present?
Mathematics
1 answer:
shutvik [7]3 years ago
6 0

Answer:

  900

Step-by-step explanation:

We assume that your 4-digit number must be in the range 1000 to 9999. Clearly, any number ending in zero will meet your requirement:

  1000/100 = 10

  3890/389 = 10

However, the requirement cannot be met when the 1s digit is other than zero.

__

For some 3-digit number N and some 1s digit x, the 4-digit number will be

  4-digit number: 10N+x

Dividing this by N will give ...

  (10N+x)/N = 10 remainder x

N will only be a factor of 10N+x when x=0.

So, there are 900 4-digit numbers that meet your requirement. They range from 1000 to 9990.

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Subtract 8 from me. Then add 21. If you divide me by 4 and add 21 you get 29. What number am I?
enyata [817]

Answer:

19

Step-by-step explanation:

x - 8 + 21 ÷ 4 + 21 = 29

x - 8 + 21 ÷ 4 = 29 - 21

                     = 8

x - 8 + 21 = 8 x 4

              = 32

x - 8 = 32 - 21

       = 11

x = 11 + 8

  = 19

Therefore the number is 19. A bit confusing to me at first but I solved it :)

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Help! What is the area of the triangle in this coordinate plane?
Novay_Z [31]

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Step-by-step explanation:

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Then divide it by 1/2.

The answer is 16.

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I don’t get this..please help!
Maslowich

Answer:

OPTION 1

Step-by-step explanation:

You just basically plug in x and y values in all the equations and check if you get the same answer each time.

Option 1 is correct because when you plug in let's say the first coordinates,10 and 4  it does give you 6 which is right.

Or you can test it the other way, just plugging in x values and see if you get the right y value for it shown above in the table.

Hope this helps!

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